I am in the following situation. I have two (rather explicit and specific) dg commutative algebras $R,S$ over a field of characteristic $0$. In fact, $S$ is an $R$-algebra, in that I have a map $R \to S$. Because I am interested in computing some derived tensor products $S \otimes_R$, I have worked out a "Koszul" resolution $\tilde S$ of $S$ over $R$. So all seems well.

But! Actually, $S$ and $R$ are both Gerstenhaber algebras (in dg vector spaces; the differential and the Gerstenhaber bracket point in opposite directions), and the map $R \to S$ is a homomorphism of Gerstenhaber algebras. My problem is that I have been so far unsuccessful at giving the resolution $\tilde S$ a Gerstenhaber structure such that the resolved map $R \to \tilde S$ is a homomorphism of Gerstenhaber algebras.

This leads me to two questions. The second question depends on the answer to the first.

Question 1: Does there necessarily exist a resolution of $S$ that computes the derived $S\otimes_R$ and that is Gerstenhaber in a compatible way?

Question 2 if the answer to 1 is YES: How do I construct it?

Question 2 if the answer to 1 is NO: Certainly my homotopy equivalence $S \leftrightarrow \tilde S$ allows me to move the Gerstenhaber structure on $S$ to something on $\tilde S$. What structure on $\tilde S$ does it move to?

the minimal model for "infinity Gerstenhaber" can be described "explicitly" since the Gerstenhaber operad is Koszul; it is generated by operations that look like symmetric products of free Lie words. In your case, since the commutative product part is formal, some part of this structure will collapse. At the very least, the "C-infinity" part, which corresponds to symmetric products of the one-term Lie word without any brackets, should be the identity in arity 1, the product in arity 2, and then 0 beyond that. I suspect the answer to question 1 is yes, which would make this comment moot.
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Gabriel C. Drummond-ColeApr 29 '11 at 6:37

As Gabriel says above a resolution can be computed. However that resolution is necessarily large and potentially unwieldy. Fortunately in many nice examples smaller resolutions exist. For example if you have a Lie-Rinehart algebra (also called a Lie-algebroid), then you may freely generate a Gerstenhaber algebra from it. If you wanted a cofibrant model of this Gerstenhaber algebra it would be enough to take a cofibrant model of the Lie-Rinehart algebra and then take the Gerstenhaber algebra of that. The resulting resolution would be far smaller.
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James GriffinApr 29 '11 at 15:03

Following on from the comment above, does your algebra S have any special properties? Is it in the image of some functor from a different category of algebras? You said that you have a "Koszul" resolution, can you put the Gerstenhaber bracket on those generators? If not, is there another resolution on which you can?
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James GriffinApr 29 '11 at 15:09

1 Answer
1

Question 1: Does there necessarily
exist a resolution of S that computes
the derived $S\otimes_R$ and that is
Gerstenhaber in a compatible way?

Yes. As pointed out in the comments, the category of dg Gerstenhaber algebra admits a model structure in which the weak equivalences are the quasi-isomorphisms, fibrations are degreewise surjections, and cobifrant obects are those dg Gerstenhaber algebras that are free as graded algerbas.

This actually wors with dg algebras over any given operad $\mathcal O$ (in place of Gerstenhaber).

Then there is also a natural model structure on the category of dg Gerstenhaber $R$-algebras (there is a more general statement about existence of a model structure on the category of objects under a given one $X$ in a model category $\mathcal C$).

So, the answer to the title of your question is that you don't "need" to know what a $G_\infty$-algebra is.

Btw, the above paper also tells you what is the definition of a $G_\infty$-algebra.

Question 2 if the answer to 1 is NO:
Certainly my homotopy equivalence $S\leftrightarrow\widetilde{S}$
allows me to move the Gerstenhaber
structure on $S$ to something on $\widetilde{S}$.
What structure on $\widetilde{S}$ does it move to?

Even though the answer to Question 1 is YES, there is still something to say here.
There is on $\widetilde{S}$ a $G_\infty$-structure. This is "just" homotopy transfer formula (and the use of the explicit minimal model for the Gerstenhaber operad).